Puzzles Archive
This is a list of the previous puzzles that have been sent out by
E-mail.
Don't forget to signup for "The MindBender" here.
To see the answer, click and hold your mouse button just to the right
of the red arrow 
and drag down. This will highlight
the answer and make it visible.
January 17, 2000
MindBender
Magic Square
-------------
| X | X | X |
-------------
| X | X | X |
-------------
| X | X | X |
-------------
Replace the Xs in the above square with the numbers from 1 to 9 so that the sums of each
row, each column, and both diagonals are the same.
___________________________
Mini-MindBender for Kids
Chess
Two grandmasters played five games of chess. Each won the same number of
games and lost the same number of games. There were no draws in any of the
games. How can this be so?
___________________________
...Answer to MindBender
Magic Square
The first step is to determine what the same total should be. The sum
of all 9 numbers is 1+2+3+4+5+6+7+8+9=9*10/2=45. That is also either
the sum of the three rows or the sum of the three columns with all 9
numbers included. Therefore, the sum of any one row or column or
diagonal is 45/3=15. Next we look at each position in the square. The
four corners are each involved in three sums (the row, the column, and
the diagonal that it is part of). The four middle-side numbers are
each involved in two sums (the row and the column that it is part of).
The center square is involved in four sums (the row, the column, and
the two diagonals that it is part of). Next we look at the nine
numbers. The following are the only possible ways for each number to
be included in a sum of 15 without any duplicate numbers used:
1: 1+9+5 1+8+6 Two ways for each odd number other than 5
3: 3+8+4 3+7+5
7: 7+6+2 7+5+3
9: 9+5+1 9+4+2
2: 2+9+4 2+8+5 2+7+6 Three ways for each even number
4: 4+9+2 4+8+3 4+6+5
6: 6+8+1 6+7+2 6+5+4
8: 8+6+1 8+5+2 8+4+3
5: 5+9+1 5+8+2 5+7+3 5+6+4 Four ways for 5
Therefore, the 5 must go in the center; the even numbers must go in
the corners; and the odd numbers (other than 5) must go in the
middle-sides. Place the 5 in the center and the 1 and 9 in two
opposite middle-sides as follows:
-------------
| X | 1 | X |
-------------
| X | 5 | X |
-------------
| X | 9 | X |
-------------
From above choose the only other addition possibility for 9 (9+4+2)
and put 4 and 2 next to the 9. Either way will end up with the same
solution because of symmetry. Your magic square will look like this:
-------------
| X | 1 | X |
-------------
| X | 5 | X |
-------------
| 4 | 9 | 2 |
-------------
The rest of the numbers almost fill themselves in!
-------------
| 8 | 1 | 6 |
-------------
| 3 | 5 | 7 |
-------------
| 4 | 9 | 2 |
-------------
The above solution is the only solution other than mirror images or
rotational variants.
This is a very old MindBender, but the solution is from the moderator.
___________________________
Answer to Mini-MindBender for Kids
Chess
They were not playing each other.
This MindBender was modified from a puzzle in Paul Sloane's book,
"Lateral Thinking Puzzlers."
January 24, 2000
MindBender
Square Square
-------------
| X | X | X |
-------------
| X | X | X |
-------------
| X | X | X |
-------------
Replace the Xs in the above square with the numbers from 1 to 9 so that
each row and the top-left to bottom-right diagonal are each perfect
squares.
___________________________
Mini-MindBender for Kids
Mrs. Jones
Mrs. Jones has two children. At least one is a boy. What are the chances
that both are boys?
___________________________
...Answer to MindBender
Square Square
The only numbers to produce 3 digit squares are 10 through 31. The squares
of 10, 11, 12, 15, 20, 21, 22, 26, and 30 have duplicate digits in them,
so they are eliminated. That leaves us with the following squares to work
with: 169, 196, 256, 289, 324, 361, 529, 576, 625, 729, 784, 841, and 961.
Now we will try each of those numbers as the top-left to bottom-right
diagonal. 625, 841, and 961 are eliminated as diagonal candidates because
there would be no other number with the same first digit to go in row one.
169 and 196 are eliminated as diagonals because either one uses up the
other digits in the other number that would have to be in row one since
row one would need to start with 1. Consider each of the remaining diagonal
possibilities. 256 is eliminated because there is no number with 5 in the
tens digit for row two. If 289 is the diagonal, 256 must be row one; 784
must be row two; row three would need to be either 139 or 319 to use up the
remaining digits 1-9 and neither are squares. Similar reasoning eliminates
all other diagonal possibilities except 324. This results in the following
unique solution:
-------------
| 3 | 6 | 1 |
-------------
| 5 | 2 | 9 |
-------------
| 7 | 8 | 4 |
-------------
I don't remember the source for this MindBender, but the solution is from
the moderator.
___________________________
Answer to Mini-MindBender for Kids
Mrs. Jones
For two children, there are only four possibilities:
.. Older . Younger
A . Boy ... Boy
B . Boy ... Girl
C . Girl .. Boy
D . Girl .. Girl
Each of these possibilities is equally likely, i.e., there is a one in
four chance that any two child family will have any one of the above
combinations. For Mrs. Jones, the possibilities are narrowed down to the
three cases -- A, B, or C. Of these, only one case -- A -- means that both
are boys. Thus, the chance that both her children are boys is one in three.
This MindBender was modified from a puzzle in Paul Sloane's book,
"Lateral Thinking Puzzlers."
January 31, 2000
MindBender
How Far
A man is riding a bike from one town to another. On the first day he
covered 1/5 of the distance. On the second day he covered 1/4 of the
remaining distance. On the third day he covered 1/3 of the distance left
to go. On the fourth day he covered 1/2 of the remaining distance. He has
now gone a total of 400 miles. How much farther is it to his destination?
___________________________
Mini-MindBender for Kids
Mrs. Brown
Mrs. Brown has two children. The younger one is a boy. What are the chances
that both are boys?
___________________________
...Answer to MindBender
How Far
There are 100 miles left to go. On day one he covers 1/5 of the total
distance. On the second day he covers 1/4 * 4/5 = 1/5 of the total
distance. On the third day he covers 1/3 * 3/5 = 1/5 of the total distance.
On the fourth day he also covers 1/5 of the total distance (1/2 * 2/5). He
has then covered 4/5 of the total distance or 400 miles, so there are 100
miles to each fifth.
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Mindbending Puzzles."
___________________________
Answer to Mini-MindBender for Kids
Mrs. Brown
For two children, there are only four possibilities:
.. Older . Younger
A . Boy ... Boy
B . Boy ... Girl
C . Girl .. Boy
D . Girl .. Girl
Each of these possibilities is equally likely, i.e., there is a one in four
chance that any two child family will have any one of the above
combinations. For Mrs. Brown, the possibilities are narrowed down to the
two cases -- A and C. Of these, only one case -- A -- means that both are
boys. There is thus a one in two chance that both her children are boys.
This MindBender was modified from a puzzle in Paul Sloane's book,
"Lateral Thinking Puzzlers."
February 2, 2000
MindBender
Bonus MindBender - Today
Today is a very special day. It is the very first day of its kind in
your lifetime, no matter how old you are. Try to figure out what
makes February 2, 2000 so special.
___________________________
...Answer to MindBender
Bonus MindBender - Today
Today, February 2, 2000, is the first day of your life where the
date is specified with all even digits (2/2/2000 or 02/02/2000).
The last such date was August 28, 888 or 8/28/888 or 08/28/0888,
over 1111 years ago.
I don't know any source for this problem.
February 7, 2000
MindBender
A Series
What are the next two numbers in the following series?
902 803 705 607 511 413 317 219 ? ?
___________________________
Mini-MindBender for Kids
Painting
A man stands in front of a painting and says, "Brothers and sisters I have
none. But this man's father is my father's son." How is the man in the
painting related to the man in front of the painting?
___________________________
...Answer to MindBender
A Series
123 29
The series is formed by adding successive primes to 900, 800, 700, etc.
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Mindbending Puzzles."
___________________________
Answer to Mini-MindBender for Kids
Painting
The man's son is in the painting. "My father's son" must be the man himself
since he has no brothers or sisters. Therefore, "this man's father is my
father's son" becomes "this man's father is I" or "I am this man's father."
So the man in the painting is his son.
This MindBender was modified from a puzzle in Paul Sloane's book,
"Lateral Thinking Puzzlers."
February 14, 2000
MindBender
Sequence Formula
What is the next number in the following sequence:
0 2 6 12 20 30 ?
Also, what is the formula for the general Nth term in the sequence?
___________________________
Mini-MindBender for Kids
Mind Reading
Pick any number. Multiply it by 2. Add 8 to the result. Divide that
result by 2. Multiply that result by 3. From that result, subtract
your original number. Subtract 7 from that result. Tell me what the
final answer is. I can now read your mind and tell you what your
original number was. Why is this possible?
___________________________
...Answer to MindBender
Sequence Formula
The next number is 42.
The Nth term in the sequence = N*N - N or N*(N-1)
This MindBender was modified from a puzzle in Jerry Stickel's
"Mindbending Puzzles" calendar for 2000.
___________________________
Answer to Mini-MindBender for Kids
Mind Reading
I simply mentally subtract 5 from your final answer and then mentally
divide that result by 2 to determine your original number, appearing
to think deeply the whole while. This works because, as simple Algebra
shows, your final result is always 2X+5 where X is your original
number. Use this to amaze your friends and family with your "mind
reading" abilities.
The MindBender moderator is the source for this Mini-MindBender.
February 21, 2000
MindBender
Cash Gift
A very wealthy man visited a high school and decided to give a
gift of cash to each upper class pupil - $40 for each junior present
and $100 for each senior present. But when he realized that there were
a total of 575 students, he realized that he did not bring enough cash
with him. He decided to wait until the night of the big football game
when 60% of the seniors would be gone. How much did it cost him?
___________________________
Mini-MindBender for Kids
900 Square
-------------
| X | X | X |
-------------
| X | X | X |
-------------
| X | X | X |
-------------
.. 9 . 0 . 0 .
Replace the Xs in the above square with the numbers from 1 to 9 so
that the sum of the three horizontal 3-digit numbers is 900.
___________________________
...Answer to MindBender
Cash Gift
You don't need to know how many seniors and juniors there are.
40% of $100 is $40 so on AVERAGE he gave each senior (even the ones
absent) $40. He also gave each junior $40. Therefore, he spent
575 * $40 = $23,000.
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Mindbending Puzzles."
___________________________
Answer to Mini-MindBender for Kids
900 Square
Since the sum of the 9 digits is 45 (9*10/2), the sum of the two
right most columns must be close to 36 (45-9) with the first carry
included. The only way this can be is if the carries are both 2. This
means the sum of the first column is 7 without the carry. The only
way this can happen is with 4, 2, and 1. The rest of the digits can
now be easily filled in. One of many possible answers is:
-------------
| 4 | 9 | 8 |
-------------
| 2 | 6 | 7 |
-------------
| 1 | 3 | 5 |
-------------
.. 9 . 0 . 0 .
This MindBender was modified from a puzzle in Raymond Blum's book,
"Math Trick, Puzzles & Games."
February 28, 2000
MindBender
3 Digit
I'm thinking of a three digit number.
The sum of the three digits equals the product of the first two
digits and also equals the product of the last two digits.
What is the number? How many such three digit numbers are there?
___________________________
Mini-MindBender for Kids
5 Numbers
I'm thinking of five numbers. The sum of these five numbers is 25.
The product of these five numbers is 945. Two of the numbers are 1
and 9. What are the five numbers?
___________________________
...Answer to MindBender
3 Digit
There are three solutions: 000, 242, and 333.
Let the first digit be F, the middle digit be M, and the last digit
be L. Since F+M+L=F*M=M*L, then F=L. If F=L=0, M=0. If F=L=1, we have
M+2=M which is impossible. If F=L=2, we have M+4=2M or M=4. If F=L=3,
we have M+6=3M or M=3. If F=L=4, we have M+8=4M, which has no integer
solution. There are also no integer solutions for F=L=5, F=L=6, F=L=7,
F=L=8, and F=L=9. Therefore, the only solutions are 000, 242, and 333.
The MindBender moderator is the source for this MindBender.
___________________________
Answer to Mini-MindBender for Kids
5 Numbers
The product of the three unknown numbers is 945/9=105. The prime
factors of 105 are 3, 5, and 7. These three numbers add to 15 which is
25-1-9. Therefore, the original five numbers are 1, 3, 5, 7, and 9.
This MindBender was modified from a puzzle in Jerry Stickel's
"Mindbending Puzzles" calendar for 2000.
March 6, 2000
MindBender
Three Times
Kris is 10 years older than her sister Katie. There was a time when
Kris was three times as old as Katie. How old was Kris when she was
three times as old as Katie? Try to solve this MindBender in more
than one way.
___________________________
Mini-MindBender for Kids
Age
A math teacher was very much involved with numbers, much like all
MindBender Solvers. He was asked by his students how old he was. He
gave them enough information to figure his age out. He said: My son
is 24 years younger than I am. He, in turn, is 25 years older than
my grandson. My grandson and I together are 73 years old. How old
is the math teacher?
___________________________
...Answer to MindBender
Three Times
Three ways (there are certainly more):
If Kris was three times as old as Katie at some time, and was then
10 years older than Katie (since the difference in their ages is
constant), 10 was twice Katie's age at that time. Therefore, at that
time, Katie was 5 and Kris was 15.
Let K be Kris' age when she was three times as old as Katie. Then
K=3(K-10) or K=3K-30 or 2K=30 or K=15.
Let K be Kris' current age. Let L be Katie's current age. Let X be
the number of years ago that Kris was three times as old as Katie.
We then have:
L=K-10
K-X=3(L-X)
This seems to be two equations in three unknowns and thus has no
unique solution. However use L=K-10 in the second equation to get:
K-X=3(K-10-X) or
K-X=3K-30-3X or
2K-2X=30 or
K-X=15 and K-X is exactly what we were to find, Kris' age when she
was three times as old as Katie.
This MindBender was modified from a puzzle in Jerry Stickel's
"Mindbending Puzzles" calendar for 2000.
___________________________
Answer to Mini-MindBender for Kids
Age
Eliminate the son from the problem and it becomes a simple problem
in two unknowns. The first two statements mean that the math teacher
is 49 years older than his grandson. Using T for the teacher's age
and G for the grandson's age, this means that T-G=49. We also know
that T+G=73. Adding the two equations results in 2T=122 or T=61.
Therefore the teacher is 61 years old.
This MindBender was modified from a puzzle in Robert M¸ller's book,
"The Great Book of Math Teasers."
March 13, 2000
MindBender
Cryptarithmetic
This is a Cryptarithmetic puzzle. It represents a long-division
problem, where each digit has been replaced by a different letter.
You are to determine the digit that corresponds to each letter.
When you are done, arranging the letters in order from the letter
corresponding to 0 up to the letter corresponding to 9 will spell
out a short English word or phrase.
... . . . DAM
.. . +-------
ANI | ARDENT
.. . . ATMD
.. . . ----
... . . RLRN
... . . RELA
... . . ----
.. . . . IDIT
.. . . . IRLM
.. . . . ----
... . . . BEN
Note: The divisor, dividend, and quotient are all English words.
___________________________
Mini-MindBender for Kids
8-Sided Dice
You roll two dice. However, these are not 6-sided dice. They are
8-sided dice that each have 1, 2, 3, 4, 5, 6, 7, and 8 dots on their
8 sides. What are the chances that you will roll an 8 (the total of
the up-faces of both dice)?
___________________________
...Answer to MindBender
Cryptarithmetic
... . . . 972
.. . +-------
781 | 759480
.. . . 7029
.. . . ----
... . . 5658
... . . 5467
... . . ----
.. . . . 1910
.. . . . 1562
.. . . . ----
... . . . 348
And the zero-to-nine word is TIMBERLAND.
There are many ways of approaching this MindBender. What I used was:
since D*I=D, A*I=A, and M*I=M, then I=1. Since N-A=1, N=A+1.
A*ANI=RELA and R is >1 since it is not 0 (first digit of RELA) and
I=1 from above. Therefore, A,N must be >= 5,6 since 4*451 is not >2000.
Trying the four possible pairs for A,N (5,6 or 6,7 or 7,8 or 8,9),
results in only one possibility: A,N = 7,8. The rest is easy.
This Mindbender came from:
http://www.cs.umbc.edu/cgi-bin/mayfield-hook/crypt
a web page that automatically generates Cryptarithmetic puzzles.
Each time you return to this page, you will get a different puzzle!
___________________________
Answer to Mini-MindBender for Kids
8-Sided Dice
The chances are 7 out of 64. Since each die has eight sides, there
are 8*8=64 total combinations of results when you roll both dice at
the same time. There are 7 ways to roll a total of 8:
(1,7) (2,6) (3,5) (4,4) (5,3) (6,2) (7,1)
so the chances are 7 out of 64.
This MindBender was modified from a puzzle in Jerry Stickel's
"Mindbending Puzzles" calendar for 2000.
March 20, 2000
MindBender
True/False
Consider the following statements about some of my kids:
A. Terri is blond.
B. Jeannie is over six feet tall.
C. Suzie is in her 27th year.
D. This statement is true if A, or B, or both A and B are true;
otherwise it is false.
E. This statement is true if B is false and vice versa.
F. This statement is false if and only if both C and E are true.
G. Just one of statements D, E, and F is true.
If I tell you that G is true, what are the true/false status of
statements A through F?
___________________________
Mini-MindBender for Kids
Seating
Some people are seated on the left and right sides of the center
aisle in a room. If one person moves from the left side to the right
side, an equal number of people are then on the left and right sides.
If that person moves back to the left side and one more person moves
from the right side to the left side, the left side has twice as
many people as the right side. How many people were originally on
the left and right sides respectively?
___________________________
...Answer to MindBender
True/False
Use a truth table for the 8 possible combinations of truth for A, B,
and C. Then fill in the resulting true/false values for statements
D, E, F, and G from the statements.
.. A B C .. D E F G
1. T T T .. T F T F
2. T T F .. T F T F
3. T F T .. T T F F
4. T F F .. T T T F
5. F T T .. T F T F
6. F T F .. T F T F
7. F F T .. F T F T
8. F F F .. F T T F
Only combination 7 has G being true so combination 7 gives the
true/false status of statements A through F.
This MindBender was modified from a puzzle in David J. Bodycombe's
book, "The Mammoth Puzzle Carnival."
___________________________
Answer to Mini-MindBender for Kids
Seating
Algebra helps us solve this problem. Let L be the number of people
originally on the left side and R be the number of people originally
on the right side. Then we have:
L-1=R+1 or L=R+2
L+1=2(R-1)
and then substituting L=R+2 into the second equation gives us:
R+2+1=2(R-1) or
R+3=2R-2 or
R=5 which then gives us L=7. So 7 people were originally on the left
side and 5 people were originally on the right side.
This MindBender was modified from a puzzle in Robert Müller's book,
"The Great Book of Math Teasers."
March 27, 2000
MindBender
Good/Bad
You come across a fork in the road. You know that one path leads you
to the land of the good, and the other leads you to the land of the
bad. However, you do not know which path leads to where. There is a
person from one of the lands standing in the middle of the fork, you
do not know if he is from the land of the good or the land of the bad.
You DO know that people from the land of the good ALWAYS tell the
truth. And the people from the land of the bad ALWAYS lie. You want
to go to the land of the good. If you are only allowed to ask one
question, how can you be certain the person will lead you to the land
of the good?
___________________________
Mini-MindBender for Kids
Box Size
A rectangular box has six sides that have surface areas of 12, 12, 24,
24, 32, and 32 square units respectively. What are the dimensions of
the box?
___________________________
...Answer to MindBender
Good/Bad
This is a classic liar/truth teller puzzle. The usual solution says
you need to ask a conditional question so that both the liar and the
truth teller answer the same. Something like "If I asked you if the
left fork was the way to the land of the good, would you say 'yes'?"
works to always get a "Yes" answer if it is the way to the land of
the good and a "No" answer if it is not.
If you ask, "Is the left fork the way to the land of the good?", and
it IS the way to the land of the good, a truth teller will say "Yes,"
and a liar will say "No." But if you ask them essentially what their
answer to the above question would be, the truth teller will still say
"Yes," but the liar will now lie about his earlier answer and say
"Yes" also. However, the danger always exists that the liar's goal is
not simply to lie but to give false information, in which case, you
can't trust his answer to such a conditional question. So this
conditional question seems to work, but not as well as the answer that
MindBender solver Stephen Tsai submitted.
The following answer is much more elegant and pleasing! And it seems
immune from the liar's desire to confuse.
"Is the left fork the way to your village?"
The truth teller will say "Yes" if it is the way to the land of the
good and "No" if it is not. The liar will answer identically.
This MindBender came from one of the MindBender solvers, Stephen Tsai.
Thank you very much, Stephen!
___________________________
Answer to Mini-MindBender for Kids
Box Size
Algebra helps us solve this problem again. Let A, B, and C be the
dimensions of the box. Then we have:
A*B=12
A*C=24
B*C=32
which gives us B=12/A from the first equation and
C=24/A from the second equation and
substituting into the third equation gives us (12/A)*(24/A)=32 or
288/(A*A)=32 or 288=32*A*A or 9=A*A or A=3.
B=12/A means B=12/3 or B=4.
C=24/A means C=24/3 or C=8 so the dimensions of the box are 3 X 4 X 8.
Trial and error will also solve this Mini-MindBender.
This MindBender was modified from a puzzle in David J. Bodycombe's
book, "The Mammoth Puzzle Carnival."
April 3, 2000
MindBender
Crazy Clock
Here's a clock puzzle, just because we in the U.S. started daylight
savings time this past weekend.
The alarm cock has gone crazy. It was correct at midnight but
immediately began to lose 12 minutes per hour. It now shows one
o'clock in the morning, but it stopped ten hours ago. What time is it?
___________________________
Mini-MindBender for Kids
Number Cross
Replace each of the question marks in the matrix below with a number,
so that each addition (+), subtraction (-), and multiplication (X)
calculation (horizontal and vertical) is correct.
... ? + ? = 10
... X . - .. +
... ? X ? = ?
.. 18 + 1 = ?
___________________________
...Answer to MindBender
Crazy Clock
If it was losing 12 minutes per hour, then it showed 48/60 or 4/5 of
the correct elapsed minutes when it stopped. One o'clock shows 60
minutes elapsed. So 60=4/5*X or X=75. So the correct time would have
been 75 minutes past midnight or 1:15. But it stopped 10 hours ago, so
now it is 11:15 in the morning.
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Astounding Puzzles."
___________________________
Answer to Mini-MindBender for Kids
Number Cross
... 6 + 4 = 10
... X . - .. +
... 3 X 3 = 9
.. 18 + 1 = 19
This MindBender was modified from a puzzle in Victor Serebriakoff's
book, "The Mammoth Book of Mindbending Puzzles."
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